If we have a subgroup H of a group G, and act by left regular action on the set A of left cosets of H in G. This induces a homomorphism $\phi: G \rightarrow S_{n}$, where ker($\phi$) is a normal subgroup of G. Is the ker($\phi$) also a subgroup of H?

  • $\begingroup$ Are you sure you mean action of $G$ on $H$? Either it's the other way around, or you're talking about the action of $G$ on the left cosets of $H$. $\endgroup$ – M. Vinay May 3 '16 at 4:03
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    $\begingroup$ If by left regular action of $G$ on $H$, you mean the element $g \in G$ acts on $h \in H$ by $h \mapsto gh$, then this not an action on (the set) $H$ (since it clearly maps elements of $H$ to elements of coset $gH$, which, in general, is not equal to $H$). $\endgroup$ – M. Vinay May 3 '16 at 4:08
  • $\begingroup$ Yes M. Vinay that is what I meant. My mistake $\endgroup$ – dpa97 May 3 '16 at 4:53

Yes. Notice that the kernel is in $H$.

If $g$ is in the kernel, we have $gH=H$. Thus $g\in H$.


Yes, $\ker \varphi \le H$. Indeed, it is the largest subgroup of $H$ that is normal in $G$.

Proof: Observe that $k \in \ker \varphi$ if and only if $kgH = gH$ for all $g \in G$, or equivalently, $g^{-1}kg \in H$ for all $g \in G$.

Taking $g = 1$, the identity element, we get $k \in H$, thus $\ker \varphi \le H$, and we know that $\ker \varphi \unlhd G$ (the notation means $\ker \varphi$ is a normal subgroup of $G$). Thus, $\ker \varphi$ is a normal subgroup of $G$ that is contained in $H$.

On the other hand, if $N$ is any normal subgroup of $G$ that is contained in $H$, then $g^{-1}kg \in N \subseteq H$, for all $g \in G$, so that $N \le \ker \varphi$, by the earlier condition. Thus, every subgroup of $H$ that is normal in $G$ is contained in $\ker \varphi$. $\square$


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